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I understand the statement for the mapping property of quotient rings but I want to know the significance of picking an ideal contained in the kernel of R.It seems that this has something to do with the fact there is a further quotient?

For example, our lecturer told us the mapping property is the reason why the following two rings are isomorphic:

  1. R[x]/<2x,3x>

where <2x,3x> denote the ideal generated by the two polynomials 2x and 3x.

  1. (R[x]/<2x>)/<3x> (there seems to be another form for this, which is (R[x]/<2x>)/<3x+<2x>>

I guess my question is why the mapping property has something to do with the fact that it makes no difference to introduce one relation at a time and to introduce two relations together?

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Define \begin{align*} \phi : R[x]/(2x)&\to R[x]/(2x,3x)\\ p(x) + (2x)&\mapsto p(x) + (2x,3x). \end{align*} It is not hard to verify that this is a ring homomorphism, and it is clearly surjective. Hence, $(R[x]/(2x))/\ker\phi\cong R[x]/(2x,3x)$ by the first isomorphism theorem.

Any element of the form $3x p(x) + (2x)$ is in the kernel, as $3x p(x)\in (2x,3x)$, so $(3x + (2x))\subseteq\ker\phi$.

Conversely, suppose $p(x) + (2x)\in\ker\phi$. Then $p(x)\in (2x,3x)$ by definition of $R[x]/(2x,3x)$ That is, $p(x) = 2x q(x) + 3x r(x)$ for some polynomials $q$ and $r$ in $R[x]$. But then, $$ p(x) + (2x) = 2x q(x) + 3x r(x) + (2x), $$ and $$ 2x q(x) + 3x r(x) + (2x) = 3x r(x) + (2x), $$ as $2x q(x) + 3x r(x) - 3x r(x) = 2x q(x)\in (2x)$. So $p(x) + (2x) = 3x r(x) + (2x)$, and hence $p(x) + (2x)\in (3x + (2x))$, so $\ker\phi\subseteq (3x + (2x))$.

Then $\ker\phi = (3x + (2x))$, which gives the isomorphism you were wondering about. (As for the "other form" of $(R[x]/(2x))/(3x + (2x))$, it technically doesn't make sense, as $3x$ is not an element of $R[x]/(2x)$, strictly speaking. I would say that $(R[x]/(2x))/(3x)$ is really just shorthand for $(R[x]/(2x))/(3x + (2x))$.)

In general, if you have an ideal $I = (a,b)\subseteq R$, then $R/I\cong (R/(a))/(b + (a))\cong (R/(b))/(a + (b))$. This is akin to saying that if you divide an element by both $a$ and $b$ (maybe pretend $R = \Bbb Z$ for a moment), the remainder is the same if you had divided first by $a$, and then by $b$ or vice versa, and this is also the same as finding the remainder when you divide by their gcd, if it exists.

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